Laplace Inverse of 1 is 1/s or Deta(t) I am confused , can anybody give me confirm answer

Thanks a lot in advance

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- Feb 19th 2013, 08:30 AMssssLaplace Inverse
Laplace Inverse of 1 is 1/s or Deta(t) I am confused , can anybody give me confirm answer

Thanks a lot in advance - Feb 19th 2013, 12:48 PMHallsofIvyRe: Laplace Inverse
Have you tried taking the Laplace transform of those?

The Laplace transform of f(x) is, by definition, $\displaystyle \int_0^\infty e^{-sx}f(x)dx$

So what are $\displaystyle \int_0^\infty (1)e^{-sx} dx$ and $\displaystyle \int_0^\infty (\delta(x))e^{-sx}dx$? - Feb 21st 2013, 06:22 AMssssRe: Laplace Inverse
I know that Laplace of 1 is 1/s but I am not getting what is Laplace Inverse of 1. Also I am not clear about answer given by you

Thanks for the reply - Feb 21st 2013, 07:06 AMHallsofIvyRe: Laplace Inverse
Surely, you understand the relation between a "Laplace transform" and its

**inverse**? (Or, for that matter, any kind of operation and it inverse!)

Yes, $\displaystyle L(1)= \int_0^\infty e^{-st}(1)dt= \left[-\frac{1}{s}e^{-st}\right]_0^\infty= \frac{1}{s}$ so 1 is the "inverse Laplace transform" of $\displaystyle \frac{1}{s}$, not 1.

The Laplace transform of [tex]\delta(x)[\tex] is $\displaystyle L(\delta(x))= \int_0^\infty e^{-st}\delta(t)dt= e^0= 1$.

Okat, NOW what is the inverse Laplace transform of "1"? - Feb 21st 2013, 07:15 AMssssRe: Laplace Inverse
According to me its Delta function